What substitution is needed to show that $\int \frac{1}{\sqrt{x^2+a^2}}\,dx=\sinh^{-1} \left( \frac{x}{a} \right) +C $?

Question

Use integration by subsitution to show that $$ \int \frac{1}{\sqrt{x^2 + a^2}}\, dx = \sinh^{-1} \left( \frac{x}{a} \right) +C $$

I’m at a loss to know what substitution to use for this integration. I tried using $u = \sqrt{x^2+a^2}$, but I just got $\ln(u)$ and got stuck from there.

I’m new to hyperbolic trig so any pointers are much appreciated as I’m used to being given what value to use for the substitution.

Answer 1

Hint: sub $$x=a\tan(\theta)\to dx=a\sec^2\theta d\theta$$ and use $$\tan^2\theta+1=\sec^2\theta$$

Answer 2

Another substituation you can use is $x=a\sinh\left(t\right)$:

then $dx=a\cosh\left(t\right)dt$ $$\begin{aligned}\int_{ }^{ }\frac{1}{\sqrt{x^{2}+a^{2}}}dx &= \int_{ }^{ }\frac{a\cosh\left(t\right)dt}{\sqrt{a^{2}\sinh^{2}\left(t\right)+a^{2}}}\\ &=\int_{ }^{ }\frac{a\cosh\left(t\right)dt}{\sqrt{a^{2}\left(\sinh^{2}\left(t\right)+1\right)}}\\ &=\frac{a}{a}\int_{ }^{ }\frac{\cosh\left(t\right)dt}{\sqrt{\left(\sinh^{2}\left(t\right)+1\right)}}\\ &=\int_{ }^{ }\frac{\cosh\left(t\right)dt}{\sqrt{\cosh^{2}\left(t\right)}}\\ &=\int_{ }^{ }dt\\ &=t+C=\sinh^{-1} \left(\frac{x}{a}\right)+C\end{aligned}$$

Also notice that $\cosh\left(t\right)$ is always strictly positive so $$\left|\cosh\left(t\right)\right|=\cosh\left(t\right)$$

Here I just used the simple identity $1+\sinh^2(t)=\cosh^2\left(t\right)$

Also, in your way, we have $\sqrt{x^{2}+a^{2}}=u$. Then $dx=\dfrac{udu}{x},$ which is not a good idea to solve this simple integral.

Answer 3

Here's one way to think about it. If $t=\tan\frac{\theta}{2}=\tanh\frac{\phi}{2}$ then$$\sin\theta=\tanh\phi=\frac{2t}{1+t^2},\,\cos\theta=\operatorname{sech}\phi=\frac{1-t^2}{1+t^2},\,\tan\theta=\sinh\phi=\frac{2t}{1-t^2}.$$This tells us that whenever you would use a sine, cosine or tangent you can instead use a hyperbolic tangent, hyperbolic secant or hyperbolic sine. In the example at hand, you probably already know how to use a tangent-based substitution, so the hyperbolic sine is natural.

Answer 4

With $u=\sqrt{x^2+a^2}$, hence $x=\sqrt{u^2-a^2}$, you get

$$\int\frac{dx}{\sqrt{x^2+a^2}}=\int\frac{u}{u\sqrt{u^2-a^2}}du=\int\frac{du}{\sqrt{u^2-a^2}},$$

not really helpful.

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If you notice the similarity between $a^2+x^2$ and $a^2-x^2$, you can try the substitution $au=ix$, and

$$\int\frac{dx}{\sqrt{a^2+x^2}}=\int\frac{a\,dx}{ia\sqrt{1-u^2}}=-i\arcsin u=\text{arsinh}\frac xa.$$

Answer 5

I think the quickest and easiest way might be to use two substitutions.

Given:

$$I = {\int}\dfrac{1}{\sqrt{x^2+a^2}}\,\mathrm{d}x$$

First, let $u=\frac{x}{a} \implies \mathrm{d}x=a\,\mathrm{d}u$.

Then you have:

$$I ={\int}\dfrac{a}{\sqrt{a^2u^2+a^2}}\,\mathrm{d}u ={\int}\dfrac{1}{\sqrt{u^2+1}}\,\mathrm{d}u$$

This where the hyperbolic substitution comes in:

$$u=\sinh\left(v\right) \implies v=\operatorname{arcsinh}\left(u\right) \implies \mathrm{d}u=\cosh\left( v \right)\,\mathrm{d}v$$

Thus we have

$$I ={\displaystyle\int }\dfrac{\cosh( v)}{\sqrt{\sinh^2(v)+1}}\,\mathrm{d}v = {\displaystyle\int }\dfrac{\cosh(v)}{\sqrt{\cosh^2(v)}}\,\mathrm{d}v = v + C= \operatorname{arcsinh}(u) + C= \operatorname{arcsinh}\left(\dfrac xa\right) + C$$